Inverse trigonometric function arise when we want to calculate the angle from side measurements in triangles. They frequently appear in the solutions of differential equations.

There different notation for the inverse function. let $\sin x$ be a function its inverse can be represented as $\sin^{-1}x$. We also write $\arcsin x$, it means the something as $\sin^{-1}x$.

Derivative of inverse trigonometric functions are given below,
$\frac{d}{dx}$$\arcsin x$ = $\frac{1}{\sqrt{1-x^{2}}}$

$\frac{d}{dx}$$\arccos x$ = $\frac{-1}{\sqrt{1-x^{2}}}$

$\frac{d}{dx}$$\arctan x$ = $\frac{1}{1+x^{2}}$

$\frac{d}{dx}$$arccot\ x$ = $\frac{-1}{1+x^{2}}$

$\frac{d}{dx}$$arcsec\ x$ = $\frac{1}{x\sqrt{x^{2}-1}}$

$\frac{d}{dx}$$arccsc\ x$ = $\frac{-1}{x\sqrt{x^{2}-1}}$

$\frac{d}{dx}$$\arcsin u$ = $\frac{1}{\sqrt{1-u^{2}}}\frac{du}{dx}$

$\frac{d}{dx}$$\arccos u$ = $\frac{-1}{\sqrt{1-u^{2}}}\frac{du}{dx}$

$\frac{d}{dx}$$\arctan u$ = $\frac{1}{1+u^{2}}\frac{du}{dx}$

Observe that the derivatives of $\arccos x$, $arccot\ x$ and $arccsc\ x$ are the negative of the derivatives of $\arcsin x$, $\arctan x$ and $arcsec\ x$ respectively.
The following problems will help us to learn more about Inverse Trig Functions

Solved Examples

Question 1: Differentiate $y$ = $\arcsin x + x\sqrt{1-x^{2}}$
Solution:
 
 $y$ = $\arcsin x + x\sqrt{1-x^{2}}$

$\frac{dy}{dx}$ =  $\frac{1}{\sqrt{1-x^{2}}}$ + x($\frac{1}{2}$)(-2x)$(1-x^{2})^{\frac{-1}{2}}$ + $\sqrt{1-x^{2}}$

= $\frac{1}{\sqrt{1-x^{2}}}$ - $\frac{x^{2}}{\sqrt{1-x^{2}}}$ + $\sqrt{1-x^{2}}$

= $\sqrt{1-x^{2}}$ + $\sqrt{1-x^{2}}$ = $2\sqrt{1-x^{2}}$

 

Question 2: Find the derivative of the function $y$ = $\arcsin (\sqrt{x})$ 
Solution:
 
$y$ = $\arcsin (\sqrt{x})$

we know that $\frac{d}{dx}$$\arcsin u$ = $\frac{1}{\sqrt{1-u^{2}}}\frac{du}{dx}$

$\frac{dy}{dx}$ = $\frac{1}{\sqrt{1-(\sqrt{x})^{2}}}\frac{d(\sqrt{x})}{dx}$

= $\frac{1}{\sqrt{1-x}}$.$\frac{1}{2\sqrt{x}}$

= $\frac{1}{2\sqrt{x-x^{2}}}$
 

Question 3: Find the derivative of the function $y$ = $(1+x^{4})\arctan (x^{2})$?
Solution:
 
$y$ = $(1+x^{4})\arctan (x^{2})$

 $\frac{dy}{dx}$ = $\arctan (x^{2})$$\frac{d}{dx}$(1+$x^{4}$) + (1+$x^{4}$)$\frac{d}{dx}$$\arctan(x^{2})$

= $\arctan (x^{2})4x^{3}$ + $(1+x^{4})$$\frac{1}{1+(x^{2})^{2}}$$2x$

= $\arctan(x^{2})4x^{3}$ + $\frac{1+x^{4}}{1+x^{4}}$$2x$

= $4x^{3} \arctan(x^{2})$ + $2X$

 

Question 4: Find the derivative of the function $y$ = $\frac{d}{dx}$$(arcsec\ e^{3x})$
Solution:
 
$y$ = $\frac{d}{dx}$$(arcsec\ e^{3x})$

we've $\frac{d}{dx}$$arcsec\ u$ = $\frac{1}{u\sqrt{u^{2}-1}}\frac{du}{dx}$

= $3 e^3x$${e^{3x}\sqrt{(e^{3x})^{2}-1}}$

= $\frac{3}{\sqrt{e^{6x}-1}}$